Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $a = \dfrac{p + 8}{-5p + 30} \times \dfrac{p^2 - 6p}{p^2 + 10p + 16} $
Explanation: First factor the quadratic. $a = \dfrac{p + 8}{-5p + 30} \times \dfrac{p^2 - 6p}{(p + 8)(p + 2)} $ Then factor out any other terms. $a = \dfrac{p + 8}{-5(p - 6)} \times \dfrac{p(p - 6)}{(p + 8)(p + 2)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (p + 8) \times p(p - 6) } { -5(p - 6) \times (p + 8)(p + 2) } $ $a = \dfrac{ p(p + 8)(p - 6)}{ -5(p - 6)(p + 8)(p + 2)} $ Notice that $(p - 6)$ and $(p + 8)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ p\cancel{(p + 8)}(p - 6)}{ -5(p - 6)\cancel{(p + 8)}(p + 2)} $ We are dividing by $p + 8$ , so $p + 8 \neq 0$ Therefore, $p \neq -8$ $a = \dfrac{ p\cancel{(p + 8)}\cancel{(p - 6)}}{ -5\cancel{(p - 6)}\cancel{(p + 8)}(p + 2)} $ We are dividing by $p - 6$ , so $p - 6 \neq 0$ Therefore, $p \neq 6$ $a = \dfrac{p}{-5(p + 2)} $ $a = \dfrac{-p}{5(p + 2)} ; \space p \neq -8 ; \space p \neq 6 $